用fortran编写·:输入一个5行5列的矩阵,完成下列要求,输出该矩阵和该矩阵的转置矩阵,求每行元素之和,将和值最大的行与第一行对调,输出对调后的新矩阵,用对角线上的各元素分别去除各元素所在的行,输出新的矩阵
时间: 2023-09-03 22:27:36 浏览: 107
### 回答1:
下面是用 Fortran 编写的代码,实现了上述功能:
```fortran
program matrix
implicit none
! 定义变量
integer :: i, j, k, temp
real, dimension(5, 5) :: a, at, b
real, dimension(5) :: rowSum, maxRowSum
! 输入矩阵
do i = 1, 5
do j = 1, 5
read(*,*) a(i, j)
end do
end do
! 计算转置矩阵
at = transpose(a)
! 输出矩阵和转置矩阵
write(*,*) "原矩阵:"
do i = 1, 5
do j = 1, 5
write(*,*) a(i, j),
end do
write(*,*)
end do
write(*,*) "转置矩阵:"
do i = 1, 5
do j = 1, 5
write(*,*) at(i, j),
end do
write(*,*)
end do
! 计算每行元素之和
rowSum = 0
do i = 1, 5
do j = 1, 5
rowSum(i) = rowSum(i) + a(i, j)
end do
end do
! 找到和值最大的行
maxRowSum = rowSum(1)
k = 1
do i = 2, 5
if (rowSum(i) > maxRowSum) then
maxRowSum = rowSum(i)
k = i
end if
end do
! 将和值最大的行与第一行对调
if (k /= 1) then
do j = 1, 5
temp = a(1, j)
a(1, j) = a(k, j)
a(k, j) = temp
end do
end if
! 输出对调后的新矩阵
write(*,*) "对调后的新矩阵:"
do i = 1, 5
do j = 1, 5
write(*,*) a(i, j),
end do
write(*,*)
end do
! 用对角线上的各元素分别去除各元素所在的行
do i = 1, 5
do j = 1, 5
if (i == j) then
b(i, j) = a(i,
### 回答2:
下面是一个使用Fortran编写的解答示例:
```fortran
program matrix_operations
implicit none
integer, dimension(5,5) :: matrix, transpose
integer :: i, j, sum_row, max_sum, max_row, temp
! 输入矩阵
write(*, *) "请输入一个5行5列的矩阵:"
do i = 1, 5
read(*, *) matrix(i, :)
end do
! 输出矩阵
write(*, *) "输入的矩阵为:"
do i = 1, 5
write(*, *) matrix(i, :)
end do
! 计算转置矩阵
do i = 1, 5
do j = 1, 5
transpose(i, j) = matrix(j, i)
end do
end do
! 输出转置矩阵
write(*, *) "转置矩阵为:"
do i = 1, 5
write(*, *) transpose(i, :)
end do
! 求每行元素之和
max_sum = sum(matrix(1, :))
max_row = 1
do i = 2, 5
sum_row = sum(matrix(i, :))
if (sum_row > max_sum) then
max_sum = sum_row
max_row = i
end if
end do
write(*, *) "每行元素之和:"
do i = 1, 5
write(*, *) sum(matrix(i, :))
end do
! 将和值最大的行与第一行对调
if (max_row > 1) then
do j = 1, 5
temp = matrix(1, j)
matrix(1, j) = matrix(max_row, j)
matrix(max_row, j) = temp
end do
end if
write(*, *) "对调后的新矩阵为:"
do i = 1, 5
write(*, *) matrix(i, :)
end do
! 对角线元素除以对应行的值
do i = 1, 5
matrix(i, i) = matrix(i, i) / matrix(i, i)
end do
write(*, *) "对角线元素除以对应行的新矩阵:"
do i = 1, 5
write(*, *) matrix(i, :)
end do
end program matrix_operations
```
这个程序首先要求用户输入一个5行5列的矩阵,然后输出该矩阵和转置矩阵。接下来,它计算每行的元素之和,并找到和值最大的行的索引,将该行与第一行对调,然后输出对调后的新矩阵。最后,它将矩阵对角线上的元素除以它们所在行的值,然后输出新的矩阵。
### 回答3:
使用Fortran编写一个程序来实现输入一个5行5列的矩阵并完成上述要求。
首先,我们声明一个5x5的矩阵和一些辅助变量:
```
program matrix_operations
implicit none
integer :: i, j
real :: matrix(5, 5), transpose_matrix(5, 5), sum_rows(5), temp
! 输入矩阵
write(*, *) "请输入一个5行5列的矩阵:"
do i = 1, 5
do j = 1, 5
read(*, *) matrix(i, j)
end do
end do
! 输出原矩阵
write(*, *) "原矩阵:"
do i = 1, 5
do j = 1, 5
write(*, "(F8.2)", advance="no") matrix(i, j)
end do
write(*, *)
end do
! 计算转置矩阵
do i = 1, 5
do j = 1, 5
transpose_matrix(i, j) = matrix(j, i)
end do
end do
! 输出转置矩阵
write(*, *) "转置矩阵:"
do i = 1, 5
do j = 1, 5
write(*, "(F8.2)", advance="no") transpose_matrix(i, j)
end do
write(*, *)
end do
! 计算每行元素之和
do i = 1, 5
sum_rows(i) = sum(matrix(i, :))
end do
! 找到和值最大的行
do i = 2, 5
if (sum_rows(i) > sum_rows(1)) then
temp = matrix(1, :)
matrix(1, :) = matrix(i, :)
matrix(i, :) = temp
end if
end do
! 输出对调后的新矩阵
write(*, *) "对调后的新矩阵:"
do i = 1, 5
do j = 1, 5
write(*, "(F8.2)", advance="no") matrix(i, j)
end do
write(*, *)
end do
! 用对角线上的元素分别除以各元素所在的行
do i = 1, 5
do j = 1, 5
if (i /= j) matrix(i, j) = matrix(i, j) / matrix(i, i)
end do
end do
! 输出新的矩阵
write(*, *) "新的矩阵:"
do i = 1, 5
do j = 1, 5
write(*, "(F8.2)", advance="no") matrix(i, j)
end do
write(*, *)
end do
end program matrix_operations
```
这个程序首先接受用户输入一个5行5列的矩阵,然后按要求输出原矩阵和转置矩阵。接下来,计算每行元素之和并将和值最大的行与第一行对调,输出对调后的新矩阵。最后,用对角线上的各元素分别去除各元素所在的行,并输出新的矩阵。
请注意,程序中的输入和输出操作可以根据需要进行更改。